WESTERGAARD COMPLEX STRESS FUNCTIONS (16)

Transcription

1 GG711c 1/1/03 1 WESTERGAARD COMPLEX STRESS FUNCTIONS (16) I Main topics A B Historical perspective Expression of the biharmonic function by harmonic functions C Boundary conditions D Use of symmetry and boundary conditions to simplify the stress function E F General solution for stresses and displacements in terms of a single complex stress function Stresses and displacements around mode I and mode II fractures II Historical perspective A B Westergaard s 1939 paper presented a simple way to express the stresses and displacements around mode I fractures (later expanded to cover mode II fractures by Sih (1966)). One of the three most important early papers in fracture mechanics, the other two being by Inglis (1913) and Griffith (191). C Westergaard s stress function is given, not derived. 1 Solution constraints a bit hard to appreciate Solution appears like magic III Expression of the biharmonic function F by harmonic functions Θ i A Harmonic functions Θ = 0 are better understood than biharmonic functions 4 F = 0 because of their greater use in many fields B All harmonic functions are biharmonic, but not all biharmonic functions are harmonic C The biharmonic function F can be expressed in terms of harmonic functions (see lecture 16 appendix and MacGregor, 1935) If Θ, Θ 0, Θ 1, Θ, Θ 3, and Θ 4 are harmonic functions (i.e., = Θi 0, then F = yθ+ Θ = xθ + Θ = x + y 0 1 Θ4 + Θ3 (16.1) Stephen Martel 16-1 University of Hawaii

2 GG711c 1/1/03 Set up of Crack Problems by Perturbation of an Ambient Stress Field Mode I σyy 00 = S σyy 00 = 0 σyy 00 = S σyy c = S σ c yy = -S (pressure) y σ c yy = 0 + x = Position of crack x=-a x=+a Crack Crack Ambient field Perturbation due to crack Total field Mode II σ yx 00 = S σ yx 00 = 0 σ yx 00 = S σ c yx = S σ c yx = -S σ c y yx = 0 x + = Position of crack x=-a Crack x=+a Crack Ambient field Perturbation due to crack Total field Mode III σ yz 00 = S σ yz 00 = 0 σ yz 00 = S σ c yz = S σ c yz = -S y σ c yz = 0 + x = Position of crack x=-a Crack x=+a Crack Ambient field Perturbation due to crack Total field Stephen Martel 16- University of Hawaii

3 GG711c 1/1/03 3 IV Boundary conditions A Mode I fracture with a uniform normal traction along y=0 σ c yy = S for x < a (16.a) By analogy with the solution for a mode III fracture, we could also imagine that this first boundary condition could be replaced by uy = CI( a z ) = CI a ( x ) for x < a, y = 0 (16.b) The other boundary conditions along the fracture plane are u y = 0 for x > a (16.3) σ yx = 0 for x a (16.4) σ yx = 0 for x > a (16.5) The second and fourth boundary conditions follow by symmetry. B Mode II fracture with a uniform normal traction along y=0 σ c yx = S for x < a (16.6a) By analogy with the solution for a mode III fracture, we can anticipate that this first boundary condition could be replaced by ux = CII( a z ) = CII a ( x ) for x < a, y = 0 (16.6b) The other boundary conditions along the fracture plane are ux = 0 for x > a (16.7) σ yy = 0 for x a (16.8) σ yy = 0 for x > a (16.9) The second and fourth boundary conditions follow by symmetry. V Use of symmetry and boundary conditions to simplify the stress function A First simplification of the stress function The plane of the crack (i.e., the plane of the boundary conditions) is y=0. An inspection of (16.1) shows that the from of the biharmonic function can be chosen as F = yθ+ Θ 0 (16.10) Our boundary conditions are along y=0, so even though both Θ and Θ 0 must be found to determine F, we can anticipate that choosing a solution in the form of (16.10) will be helpful for finding F. Stephen Martel 16-3 University of Hawaii

4 GG711c 1/1/03 4 B Second simplification of the stress function We need to know what Θ and Θ 0 are and how they are related. We use the boundary conditions to do this. This will lead us to expressions for stresses (and displacements) in terms of regular derivatives of a complex function (i.e., its real and imaginary parts solve the Laplace equation) instead of partial derivatives of a biharmonic function. We start by expressing the stresses in terms of the biharmonic function F F F F σ xx = σ yy = σ y yx = x xy (16.11) The first partial derivatives of F with respect to x and y are, respectively: F ( yθ+ Θ = 0) Θ Θ = y + Θ + 0 Θ Θ = y + 0 x x x (16.1a) F ( yθ+ Θ = 0) Θ Θ = y + Θ + 0 Θ Θ = y + Θ + 0 y y y (16.1b) To get the second partial derivatives we will make some substitutions. Let us refer to the partial derivatives in (16.1) by the following terms: Φ = Θ Ψ = Θ ψ = Θ 0 χ = Θ 0 (16.13a) In light of (16.3a), and because Θ and Θ 0 are harmonic functions Ψ ψ χ = = and (16.13b) Θ Θ Ψ = = Θ 0 Θ χ ψ y and = 0 = y (16.13c) So (16.1) and (16.13) yield F F = yψ + χ (16.14a) = yφ+ Θ+ ψ (16.14b) Now we return to the stresses. Using (16.1) and (16.13) F y ψ ψ χ σ xx = = ( Φ+ Θ+ ) Φ Φ = y + Φ + Φ + = Φ + y y y x (16.15a) F ( yψ + χ) Ψ χ Ψ χ χ σ yy = = = y + Ψ + = y + = y + x x x x x (16.15b) F y χ χ χ σ yx = = ( Ψ + ) Ψ Φ = y Ψ + = Ψ y y x y (16.15c) Stephen Martel 16-4 University of Hawaii

5 GG711c 1/1/ Mode I constraints and mode I stresses The mode I boundary conditions allow (16.15) to be simplified. For mode I, along y=0 the shear stress is zero, so from (16.15c) Ψ = χ ψ or using (16.13b) Ψ = (16.16) This relates Θ and Θ 0 for mode I. Substituting this back into (16.15c), the shear stress becomes a function of only y and Φ. σ yx = y x (16.17) Now we turn our attention to the normal stresses in (16.15). We let the derivatives of Θ be the real and imaginary parts of the following complex function; the imaginary part can be expressed using (16.16) for Ψ. χ Z I ( ζ )= Φ+ iψ= Φ+ i (16.18) For this to be worthy of consideration as a stress function (i.e., for it to have derivatives that are defined) the two Cauchy-Riemann conditions (15.0) must apply. The first Cauchy-Riemann condition, applied to (16.18), yields Ψ = (16.19a) An inspection of (16.13b) shows that Φ and Ψ always satisfy this. Applying the second Cauchy-Riemann condition to (16.18), requires that χ Ψ = χ χ = = = or using (16.16 to express Ψ) x y (16.19b) The righthand part of (16.19b) is satisfied if Φ = χ (16.0) This key constraint results from symmetry across the fracture plane. To obtain the mode I stresses we now substitute (16.0) into (16.15a) χ σ xx = Φ + y = Φ + y Φ = Φ + y y x y y, (16.a) then substitute (16.0) into (16.15a), and apply the left side of (16.19b) Ψ χ Ψ σ yy = y + = Φ + y = Φ y x x x x (16.b) and finally reproduce (16.17) σ yx = y x (16.c) Only y and the partial derivatives of Θ are used here (see (16.10)). Stephen Martel 16-5 University of Hawaii

6 GG711c 1/1/03 6 Mode II constraints and mode II stresses For mode II, the normal traction on the plane y=0 is zero, so (16.15b) gives χ = 0 (16.3) The simplest way to enforce this condition is to set χ = 0. So (16.15), when specialized for mode II, becomes σ xx = Φ + y y (16.4a) Ψ σ yy = y x (16.4b) Ψ χ Ψ σ yx = Ψ y = Ψ y = Ψ y y y y x (16.4c) 3 Closing comments on this section So the functions Φ and Ψ, along y = 0, yield the boundary conditions on the fracture. MacGregor (1935) notes that these quantities are proportional to the dilation (volumetric strain) and rotation in the body. We have shown that the stress can be obtained by two harmonic functions that are partial derivatives of the same higher order function. Similarly, these two harmonic functions can also be taken to be the real and imaginary components of a single complex function. Stephen Martel 16-6 University of Hawaii

7 GG711c 1/1/03 7 VI General solution for stresses and displacements in terms of a single complex stress function A Road map Z The work so far yields stresses in terms of the partial derivatives of some harmonic functions with respect to x and y. The associated stress functions (and stresses and displacements) can be cast in terms of the derivatives of a single complex function in terms of the complex variable ζ, as (16.18 suggests). The following functions are needed: ZZZ,,, and Z, where dz = Z Z = d ζ dz = d ζ dz dζ These functions can be found directly from the boundary conditions, but owing to time constraints we will not do this. Instead we will rely upon analogies with the solutions that we have derived so far and show that the solutions for the stress functions satisfy the boundary conditions. The stresses will come from Z and Z. The strains are proportional to the stresses, and the displacements come from integration of the strains, so the displacements must depend on the integrals of those terms. From our previous results we know that the relative displacements along a fracture vary as (a -x ) 1/, provided that the stresses promoting relative displacement of the fracture walls are uniform. We thus can anticipate a term like this being in the complex functions. Stephen Martel 16-7 University of Hawaii

8 GG711c 1/1/03 8 B Derivatives of a complex function We apply the Cauchy-Riemann conditions (see (15.8) and (15.9)) several times in taking derivatives of complex functions with respect to y: dz Re Z ImZ ImZ Re Z = Z = + i = i = Re Z + iimz dζ (16.5) C Mode I Consider a candidate mode I stress function to have the form of (16.10) F I = Re Z I + yimz I (16.6) The first derivative of F I with respect to y is: F I Z I Z y I y Z I Z = ZI y I ZI Re Im + + = Re Im Im + + Im (16.7) Using the Cauchy-Riemann relations, this can be rewritten F I = ( ImZ I )+ yre Z I + ImZ I = yre Z I (16.8) And then, noting that (16.8) feeds into (16.9a) and 16.9b), σ FI y ZI xx y ZI y = = ( Re ) Re = + Re ZI = Re ZI yimzi (16.9a) F σ I ( Re Z I + yimz I ) yy = = = Re Z I + yimz I (16.9b) FI F σ I y y ZI xy = = = ( Re ) = yre ZI xy (16.9c) Comparing (16.9) with (16.) term by term σ xx = Φ + y y, (16.a) σ yy = Φ y x (16.b) σ yx = y x (16.c) shows that Φ = ReZ I and Ψ = yrez I in the expression below (16.18) χ Z I ( ζ )= Φ+ iψ= Φ+ i (16.18) So the harmonic functions of section V are represented in the complex function Z I. Stephen Martel 16-8 University of Hawaii

9 GG711c 1/1/03 9 The stress function for mode I The elliptical relative displacement profile across a mode III fracture are given by the stress function of (15.5) ZIII = S ζ a (15.5) A mode I fracture has an elliptical relative displacement profile too (14.5), so by analogy we try the above stress function and its derivatives ZI = S ζ a (16.30a) dz ZI = I = S a S a d 1 1 / 1 / ( ζ ) ζ = ζ( ζ ) ζ (16.30b) 1 / d ( ζ a ) dz d Z I I = = 1 / ( ζ ) S ζ + ( ζ a ) dζ dζ dζ (16.30c) / / / / = ( ) ( )+ ( ) = ( 3 1 S ζ ζ a ζ ζ a S ζ ζ a ) + ( ζ a ) Substituting (16.30) into (16.9) gives the stresses, here in tri-polar form r σ θ θ + θ ya θ + θ xx = S cos 1 sin / ( r r ) ( r r ) / (16.31a) r σ θ θ + θ ya θ + θ yy = S cos 1 + sin / ( r r ) ( r r ) / (16.31b) ya θ + θ σ xy = S cos 3 / ( r r ) (16.31c) Stephen Martel 16-9 University of Hawaii

10 GG711c 1/1/03 10 Do these recover our boundary conditions? For y=0, 0<x<a x π σ yy = S + 0 cos 0 ( a x a+ x ) = 0 ( )( ) 1 / σ xy = S[]= 0 0 For y=0, -a<x<0 x σyy = S π π + 0 cos ( a x a+ x ) = 0 ( )( ) 1 / σ xy = S[]= 0 0 (16.3a) (16.3b) (16.3c) (16.3d) As r, r 1, r, and θ θ 1, θ 1 σxx S cos() 0 0sin( 3θ) S 1 = (16.33a) 1 σxx S cos() 0 0sin( 3θ) S 1 = (16.33b) σxy = S[ () 0 cos( 3θ) ]= 0 (16.33c) So the solution is for a plate under biaxial tension of magnitude S with a crack. A uniaxial compression parallel to the crack (σ xx = -S) can be added to yield the solution for a traction-free crack under a unaxial tension perpendicular to the crack. This corresponds to the right column of page 16-. To obtain the solution for the central columns of page 16-, and biaxial stress of magnitude -S must be added to the solutions of (16.31). The associated stress function and relevant derivatives are Z I = S a ζ 1 / ζ (16.34a) Z I = S a ζζ 1 / 1 (16.34b) Z S a a 3 / I = 1 3 ζ ζ (16.34c) Stephen Martel University of Hawaii

11 GG711c 1/1/03 11 Westergaard (1939) gives the following expression for the displacements (and MacGregor (1935) shows where they come from) for mode I, where G is the shear modulus: Gux = ( 1 ν )ReZI yimzi (16.35a) Gu y = 1 ( ν)imz I yre Z I (16.35a) Substituting (16.34)) these expressions into (16.3) gives the displacements. Segall and Pollard give the results. A point Segall and Pollard highlight is that geologists and geophysicists commonly are the most interested in the displacements associated with the relative motion of the fracture walls (e.g., during an earthquake), and not the displacements due to the far-field stresses. For mode II (Tada et al., 1976) ΦII = -yrez (16.36) σ xx = ImZ+ yrez (16.37a) σ yy = yre Z (16.37b) σ xy = Re Z yimz (16.37c) κ ux = + 5 ImZ + yre Z 8G (16.38a) 1 uy = κ Re Z y Im Z 4G (16.38b) Stephen Martel University of Hawaii

12 GG711c 1/1/03 1 References Griffith, A.A., The phenomena of rupture and flow in solids, Transactions of the Royal Society, v, A1, p Inglis, C.E., 1913, Stress in a plate due to the presence of cracks and sharp corners, Transactions of the Institute for Naval Architecture, v. 44, p MacGregor, C.W., 1935, The potential function method for the solution of twodimensional stress problems: Transactions of the American Mathematical Society, v. 38, p Pollard, D.D., and Segall, P., 1987, Theoretical displacements and stresses near fractures in rock>with applications to faults, joints, veins, dikes, and solution surfaces: in Atkinson, B.K., Fracture Mechanics of Rock, Academic Press, London, p Tada, H., Paris, P.C., and Irwin, G.R., 1973, The stress analysis of cracks handbook: Del research Corporation, Hellertown, Pennsylvania. Westergaard, H.M., 1939, Bearing pressures and cracks: Journal of Applied mechanics, v. 6, p. A49-A53. Stephen Martel 16-1 University of Hawaii

13 GG711c 1/1/03 13 Comparison of solutions of MacGregor, Westergaard, Tada et al., and Pollard and Segall Airy stress function MacGregor Westergaard Tada et al. P & S F F Φ U σxx σyy σxy χ Φ + y + y x χ y y x χ Ψ y + x x ReφI + y Im φi + φii [ ] [ ] [ ] Re φi + φii y Im φi + φii [ ] [ ] Im φii y Re φi + φii µux Ω 1 ( ν)imφ 1 ( ν) Ω y I y + ( 1 ν)imφii Θ 1 ( ν) Ω y y Re[ φi + φii ] x µux µuy 1 ( ν)reφii + ( 1 ν)reφi y Im[ φi + φii ] µuy Stephen Martel University of Hawaii

14 GG711c 1/1/03 14 Airy stress function I σxx σyy σxy µux µux µuy µuy MacGregor Westergaard Tada et al. P & S F F Φ 1 Re zφ χ Φ = ImW Re Z Re Z Re Z Re Z ImZ ImZ Ψ = ReW ImZ ImZ Ω Re Z Re Z Ω ImZ ImZ Θ ImZ ImZ Θ Re Z Re Z Θ 0 Re Z Re Z ReW ImZ ImW Re Z Re W Im Z Im W Re Z Re Z yimz Φ + y y Re Z yimz ReφI + y Imφ ImW yrew Re Z yimz Re Z yimz ReφI + y Imφ Re Z+ yimz Re Z+ yimz Reφ Φ y I y ImφI y ImW + yrew Re Z+ yimz Re Z+ yimz ReφI y ImφI yre Z yre Z y Reφ y I x yim W yre Z yre Z y ReφI Ω ( 1 ν ) Ω y ( 1 ν )ReZ ( 1 ν )ReZ 1 ( ν)imφi y yimz yimz y ReφI Ω ( 1 ν ) Ω y ( 1 ν )ReZ ( 1 ν )ReZ 1 ( ν)imφi y yimz yimz y ReφI Θ 1 ( ν) Θ y 1 ( ν)imz 1 ( ν)imz ( 1 ν)reφi y yre Z yre Z y ImφI Θ 1 ( ν) Θ y 1 ( ν)imz 1 ( ν)imz ( 1 ν)reφi y yre Z yre Z y ImφI Stephen Martel University of Hawaii

15 GG711c 1/1/03 15 Mode II σxx σyy σxy µux µux µuy µuy MacGregor Westergaard Tada et al. P & S Φ = ImW ImZ ImZ Im Z Im Z Re Z Re Z Ψ = ReW Re Z Re Z Ω ImZ ImZ Ω Re Z Re Z Θ Re Z Re Z Θ ImZ ImZ Θ 0 ReW Re Z Re Z ImW ImZ ImZ Re W Re Z Re Z Im W ImZ ImZ ImZ+ yrez Φ + y y y Im φii ImW yrew ImZ+ yrez y Im φii yre Z Reφ y II y y ImφII yre W yre Z ReφII y ImφII Re Z yimz Imφ Ψ y II x y ReφII ReW + yimw Re Z yimz ImφII y ReφII Ω 1 ( ν) Ω y 1 ( ν)imz ( 1 ν)imφii y + yre Z y ReφII Ω 1 ( ν) Ω y 1 ( ν)imz ( 1 ν)imφii y + yre Z y ReφII Θ ( 1 ν ) Θ y ( 1 ν )ReZ 1 ( ν)reφii y yimz y ImφII Θ ( 1 ν ) Θ y ( 1 ν )ReZ 1 ( ν)reφii y yimz y ImφII Stephen Martel University of Hawaii

16 GG711c 1/1/03 16 Mode III MacGregor Westergaard Tada et al. P & S Stephen Martel University of Hawaii

17 GG711c 1/1/03 17 Appendix for Lecture 16 I Main topic: Solution of biharmonic functions in terms of harmonic functions (see MacGregor, 1935, Timoshenko and Goodier, 1970, p. 17) Consider the harmonic functions Θ = Θ(x,y) and Θ 0 = Θ 0 (x,y). We want to show that yθ + Θ 0 is biharmonic. Clearly if Θ 0 = 0 then 4 Θ 0 = 0 (16.A1) (16.A) So the crux of the issue is yθ. ( y Θ ) y Θ y y Θ = + Θ = x x (16.A3) Θ y ( yθ) x Θ Θ y Θ = = y + = y (16.A4) Similarly, y Θ y Θ y y Θ Θ Θ y y ( ) = + = + (16.A5) Θ y ( yθ) y Θ Θ Θ Θ Θ = = y + + = y + So ( )= + Θ Θ y y + Θ Θ y The term in parentheses on the right side of the above equation is zero because Θ is harmonic (i.e., Θ = 0 ), so Θ ( yθ)= This means 4 ( )= Θ y Θ y Taking the second partial derivatives of Θ yields Θ Θ = (16.A6) (16.A7) (16.A8) (16.A9) (16.A10) Stephen Martel University of Hawaii

18 GG711c 1/1/03 18 Θ 3 Θ = 3 and, Θ Θ = x y φ 3 φ = y x Using these results ( )= 4 Θ 3 Θ 3 Θ Θ Θ yθ = + = + = 0 yy yx So yθ is biharmonic if Θ is harmonic. (16.A11) (16.A1) (16.A13) (16.A14) You might wonder why one would say if Θ is harmonic then yθ +Θ 0 is biharmonic why bother including Θ 0? We can draw an analogy with polynomial functions of a single variable. Consider the fourth derivative of f=y 3 with respect to y d 4 y 3 d 3 3 y d 6 y 1 d () 6 = = = = 0 dy 4 dy 3 dy dy (16.A15) If we integrated the second derivative (f = 6y) twice to recover a general d 4 f solution for dy 4, we would not get y 3 we would get y 3 + (Ay + B), with the term in brackets being a function of integration. Similarly, even though Θ Θ = 0, one can not just integrate ( y Θ)= y twice to obtain a a general solution. A harmonic function needs to be added to obtain a general solution for 4 f( x, y) = 0. Stephen Martel University of Hawaii